First note, by [1]}, we have that \(\operatorname{Supp}_R(M\otimes _R{{}^{\varphi ^n}\!}R)=\operatorname{Supp}_R(M)\) . In particular, \({{}^{\varphi ^n}\!}R\)  satisfies in the Nakayama's property; see Remark REF . Also, we have \(\operatorname{\textnormal {grade}}_R(\operatorname{\mathcal {Z}},{{}^{\varphi ^n}\!}R)=\operatorname{\textnormal {grade}}_R(\operatorname{\mathcal {Z}},R)\) , and that \({{}^{\varphi ^n}\!}R\)  is Tor-rigid.
(i). This follows from Lemma REF (c) and Remark REF .
(ii) and (iii). Assume that \(\operatorname{H}^i_{\operatorname{\mathcal {Z}}}(M\otimes _R{{}^{\varphi ^n}\!}R)=0\)  for some \(i\)  with \(0\le i\le \operatorname{\textnormal {grade}}_R(\operatorname{\mathcal {Z}},M)\) . Then, by Theorem REF  and Remark REF , we have
\(\operatorname{Tor}_j^R(M,{{}^{\varphi ^n}\!}R)=0\)  for all \(j\ge 1\) . In view of [2]}, \(\operatorname{pd}_R(M)<\infty \) , as claimed by (iii). Now, we proceed to give a proof for part (ii).
We are going to reduce to the \(F\) -finite case. To this end, we pass to the local ring \(S\)  that follows from Remark REF  and, without losing of generality, we may assume that \(R\)  is \(F\) -finite.
We know, by [1]}, the following equality holds for each \(\mathfrak {p}\in \operatorname{Spec}(R)\) :
\(\operatorname{pd}_{R_\mathfrak {p}}(M_\mathfrak {p})=\operatorname{pd}_{R_{\mathfrak {p}}}(M_\mathfrak {p}\otimes _{R_\mathfrak {p}}{{}^{\varphi ^n}\!}R_\mathfrak {p}).\qquad \mathrm {(\ref {fr}.2)}\) 
Therefore, it follows from (REF .2) and Proposition REF  that \(\operatorname{\textnormal {grade}}_R(\operatorname{\mathcal {Z}},M)=\operatorname{\textnormal {grade}}_R(\operatorname{\mathcal {Z}},M\otimes _R{{}^{\varphi ^n}\!}R)\)  and so the assertion follows.
(iv). This is clear by parts (ii) and (iii).
